Altitude

In triangle ABC, BC=13, CA=14, and AB=15. If D is a point on \overline{CA} such that \overline{BD} is perpendicular to \overline{CA}, what is BD?
image
Source: NCTM Mathematics Teacher, November 2006

SOLUTION
Area of triangle ABC by Heron’s formula
s=(1/2)(13+14+15)=21
Area of triangle ABC=\sqrt{21(21-13)(21-14)(21-15)}
=\sqrt{7056}
=84
Area of triangle ABC using base AC and altitude BD
(1/2)14\times BD=84
BD=12

Answer: 12

Alternative solution 1
image
Use the Pythagorean theorem in right triangle CDB
x^2+h^2=13^2=169\qquad\qquad (1)
Similarly, in right triangle ADB
(14-x)^2+h^2=15^2
196-28x+x^2+h^2=225
x^2+h^2=29+28x
From Eq. (1)
169=29+28x
x=5
Substitute the value of x into Eq. (1)
5^2+h^2=169
h^2=144
h=12

Alternative solution 2
image
Use the Pythagorean theorem in right triangles ADB and CDB
DA^2+BD^2=15^2=225\qquad\qquad (2)
CD^2+BD^2=13^2=169\qquad\qquad (3)
Subtract Eq. (3) from Eq. (2)
DA^2-CD^2=56
(DA+CD)(DA-CD)=56
AC(DA-CD)=56
14(DA-CD)=56
DA-CD=4\qquad\qquad\;\;(4)
DA+CD=14\qquad\qquad (5)
Add Eq. (4) and Eq. (5)
2DA=18
DA=9
Substitute the value of DA into Eq. (2)
9^2+BD^2=225
BD^2=144
BD=12

Alternative solution 3
image
Use the law of Cosine in triangle ABC
15^2=13^2+14^2-2(13)(14)\text{cos}\,C
225=169+196-2(182)\text{cos}\,C
225=365-364\,\text{cos}\,C
\textrm{-}140=\textrm{-}364\,\text{cos}\,C
\text{cos}\,C=140/364
In right triangle CDB
\text{cos}\,C=CD/BC=CD/13
140/364=CD/13
CD=5
Use the Pythagorean theorem in right triangle CDB
CD^2+BD^2=BC^2
5^2+BD^2=13^2
BD^2=144
BD=12

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About mvtrinh

Retired high school math teacher.
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